The trace theorem
W
p2,1(Ω
T) ∋ f 7→ ∇
xf ∈ W
1−1/p,1/2−1/2pp
( ∂ Ω
T)
revisited
Peter Weidemaier
Abstract. Filling a possible gap in the literature, we give a complete and readable proof of this trace theorem, which also shows that the imbedding constant is uniformly bounded forT↓0. The proof is based on a version of Hardy’s inequality (cp. Appendix).
Keywords: trace theory, anisotropic Sobolev spaces Classification: 46E35
Introduction.
The imbedding theorem described in the title can be found in LADYSHENS- KAYA et al. [L/S/U, Chapter II, Lemma 3.4]. However, none of the references cited there seems to contain a complete proof. The theorem is also stated in IL’IN [I, Theorem 8.4]; but there too, no proof is given. Things look even worse, if we ask for the dependence of the imbedding constantc(T) on the height T of the space- time cylinder (for smallT). In some applications of this trace theorem to nonlinear problems, one needsc(T)≤c0 for allT small (cf. WEIDEMAIER [W], particularly the Appendix). However, the formulation in IL’IN [I, Theorem 8.4], exhibits an explosion of c(T) forT ↓0. To settle things, we shall give in this note a detailed proof for the imbedding, which also shows the uniformity ofc(T) forT ↓0.
The paper is organized as follows: in Chapter 1 we deduce an integral representa- tion for∇xf in terms of∂tf, ∂x2f, which is the basis for the estimates in Chapter 2.
Let us fix the notation: ΩT := Ω×(0, T) with the typical point (x, t)∈ΩT; here Ω⊂IRn. The prime characterizes (n−1)-dimensional quantities : thus we writex∈ IRnasx= (x′, xn), x′∈IRn−1;Qn−1(a′, b′) is the open parallelepipedQn−1
j=1(ai, bi), when a′ = (a1, . . . , an−1), b′ = (b1, . . . , bn−1);Qn−1(λ) := Qn−1(−λ1′, λ1′) for λ∈IR; here
1′ := (1,· · ·,1) ∈ INn−1;Q+n(λ) := Qn−1(λ)×( 0, λ) ; the superscript ˇ always indicates the deletion of a coordinate (the n-th. one, if not further specified) , e.g.
ˇi
y= (y1,· · ·, yi−1, yi+1,· · · , yn) (1≤i≤n) and ˇQn+1(a, b) :=Qn+1
i6=ni=1
(ai, bi) . Wp2,1(ΩT) := {u|∂xαu, ∂tu (distr. sense) ∈ Lp(ΩT) ∀ |α| ≤ 2} with the obvious norm.
I thank Prof. V.A. Solonnikov, Leningrad, for valuable hints
For a bounded domain Ω ⊂ IRn, ∂Ω ∈ C2 means that ∂Ω is a C2-hypersurface.
The spaces Wpα,β(∂ΩT) (α, β ∈ (0,1)) are defined as usual, via a partition of unity on ∂Ω, and using local charts. We use the notation c∗ to emphasize the non-dependence of the constantc on the quantityT (forT small).
1. Integral representation.
Our starting point is an integral representation for ∂νf in terms of f: if f is smooth and defined on Qn−1(−λ1′,2λ1′)×[ 0,2λ]×[ 0,3T], then we have (cf.
IL’IN/ SOLONNIKOV [I/S, p. 70, (6)] withmi= 0, ki =li)
∂νf(x, t) = A Tr
Z . . .
Z
Qn+1(0, Tκ)
f((x, t) +y)Π(y, T)dy+
+
n+1
X
i=1
Bi Z T
0
v−(1+r) Z
. . . Z
Qn+1(0, vκ)
f((x, t) +y)Πi(y, v)∂ˇi iliψi(yi, v)dy dv
for (x, t)∈Q+n(λ)×[ 0, T], T ≤T0(λ) andνj ≤lj−1, where (cp. [I/S, pp. 69–70])
Π(y, T) :=
n+1
Y
j=1
∂jljχj(yj, T)
χj(yj, T) :=yljj−νj−1 Z Tκj
yj
(Tκj−s)µjsλjds ,
Πi(y, v) :=iˇ
n+1
Y
j=1 j6=i
∂jljχj(yj, v),
ψi(yi, v) :=yili+λi−νi·(vκi−yi)µi
with certain parametersµj, λj ∈IN and certainA, Bi∈IR; hereTκ:=
(Tκ1,· · · , Tκn+1), r:=κ·(1 +λ+µ), 1 := (1,· · · ,1)∈INn+1.
In the sequel we fixl:= (2,· · · ,2,1)∈INn+1, κ= (κ′, κn, κn+1) := 1l = (12,· · · ,12,1) and choose the parametersµj, λjso large that∂jkψj(yj, v) vanishes foryj= 0, yj= Tκj, 1≤k≤lj. Hence, integrating by parts and introducing Ki(y, v) := Πi(y, v)ˇi ψi(yi, v) (0≤yi≤vκi) , we have shown that
(1.1) ∂νf(x, t) = A Tr
Z . . .
Z
Qn+1(0, Tκ)
f((x, t) +y)Π(y, T)dy+
+
n+1
X
i=1
B˜i Z T
0
v−(1+r) Z
. . . Z
Qn+1(0, vκ)
∂ilif((x, t) +y)Ki(y, v)dy dv.
The kernels Π, Kiin this representation satisfy (uniformly w.r.t. y∈Qn+1(0, vκ))
|∂yαΠ(y, v)| ≤c·vr−κ·(1+ν+α) ∀ |α| ≤2 (1.2)
|∂n+1s Ki(y, v)| ≤c·ynε·vr+1−κ·(1+ν)−εκn−s
(1.3)
(∂n+1 :=∂yn+1,0≤s≤1, 1≤i≤n+ 1, ε∈[0,1)).
For the proof of these two inequalities, we first note that∂jlj+αjχj(yj, v) is a linear combination of terms of the form (vκj−yj)ρ1yρj2withρ1+ρ2=µj+λj−νj−αj, ρ2>
0 (forλj large) and consequently
|∂jlj+αjχj(yj, v)| ≤c·yεj·v−κj(ε+αj)·vκj(µj+λj−νj) (0≤yj ≤vκj) for arbitraryε∈[0,1[ ; this implies (for 1≤k≤n−1)
|∂n+1s Πk(ky, v)| ≤ˇ c·yεn·v−κnε−κn+1·s·vκ(µ+λ−ν)−κkδk
|∂n+1s Πn(ny, v)| ≤ˇ c·v−κn+1·s·vκ·(µ+λ−ν)−κnδn
|Πn+1(n+1y , v)| ≤ˇ c·yεn·v−κn·ε·vκ·(µ+λ−ν)−κn+1δn+1, whereδj:=µj+λj−νj. The definition ofψi easily implies
|ψk(yk, v)| ≤c·vκk·(lk+δk)
|ψn(yn, v)| ≤c·yεn·v−κn·ε·vκn·(ln+δn)
|∂n+1s ψn+1(yn+1, v)| ≤c·v−s·κn+1·vκn+1·(ln+1+δn+1) ;
since Ki(y, v) = Πi(y, v)ψˇi i(yi, v), κili = 1 (1 ≤ i ≤ n+ 1), κn+1 = 1, r = κ·(1 +λ+µ), these formulas yield (1.3). For (1.2) compare IL’IN/ SOLONNIKOV [I/S, p. 72].
2. Estimates.
Our aim in this chapter is to prove the imbedding Wp2,1(ΩT) ∋ f 7→ ∇xf ∈ W1−
1
p,12(1−1p)
p (∂ΩT) with the imbedding constantc∗independent ofT (forT small);
here we let Ω be a bounded domain in IRnwith boundary of the classC2. Flattening the boundary locally, it is no restriction to assume that Ω is a cube i.e. Ω =Qn+(λ).
SinceC2(Q+n(λ)×[0, T]) is dense inWp2,1(Qn+(λ)×(0, T)) (cf. R ´AKOSN´IK [R, The- orem 3]) and since the Hestenes-Whitney extension method (cf. ADAMS [A, p. 83]) yields a linear continuous extension operator
ET : Wp2,1(Qn+(λ)×(0, T))→Wp2,1(Qn+(2λ)×(0,2T)) with ET(C2(Q+n(λ)×[0, T]))⊂C2(Qn+(2λ)×[0,2T]) and
kETkW2,1
p (Qn+(λ)×(0,T))→Wp2,1(Qn+(2λ)×(0,2T)) ≤B∗ uniformly for all small T, it is sufficient to prove
k∇xfk
W1−
1p ,1 2 (1−1
p)
p (Qn−1(λ)×(0,T))
≤c∗· kfkW2,1
p (Qn+(2λ)×(0,2T))
for allf ∈C2(Qn+(2λ)×[ 0,2T]). The most difficult part in this inequality is the estimate for the time-regularity of the trace, i.e.
(2.1) |∇xf|
L0,
12 (1−1 p)
p (Qn−1(λ)×(0,T))
≤c∗· kfkW2,1
p (Qn+(2λ)×(0,2T)),
where|g|p
L0,βp (Qn−1(λ)×( 0,T)):=
T
R
0
h−(1+pβ)k∆t,hgkpp,Qn−1(λ)×( 0,T−h)dh
forβ∈( 0,1 ), when (∆t,hg)(x′, t) :=g(x′, t+h)−g(x′, t) andk · kp, X :=k · kLp(X). The estimate for the spatial regularity follows from the more elementary trace the- oremWp1(Ω)→W1−
1
p p(∂Ω) (cp. KUFNER et al. [K/J/F, 6.8.13 Theorem, p. 337]) by an easy scaling argument (in t). In the sequel, we shall prove (2.1). For this purpose, we start from the representation (1.1) for ∂jf (1 ≤ j ≤ n): splitting RT
0 (· · ·)dv =Rh
0 (· · ·)dv+RT
h (· · ·)dv in the sum in the second line in (1.1) we get
∂jf(·) =H1(·) +
n+1
X
i=1
B˜i{H2(i)(·) +H3(i)(·)}, where
(2.2.)
H1(·) := A Tr
Z . . .
Z
Qn+1(0, Tκ)
f(·+y)Π(y, T)dy,
H2(i)(·) :=
Z h
0 v−(1+r) Z
. . . Z
Qn+1(0, vκ)
∂ilif(·+y)·Ki(y, v)dy dv,
H3(i)(·) :=
Z T h
v−(1+r) Z
. . . Z
Qn+1(0, vκ)
∂ilif(·+y)·Ki(y, v)dy dv.
In the sequel, we set (γH1)(x′, t) :=H1(x′,0, t) ; we find
(2.3) k∆t,h(γH1)kp,Qn−1(λ)×(0,T−h) ≤ h· k∂t(γH1)kp,Qn−1(λ)×(0,T)
(use|∆t,hf(τ)| ≤Rh
0 |f′(τ+s)|dsand Minkowski’s integral inequality (cp. WHEE- DEN/ ZYGMUND [W/Z, p. 143])); now
|∂t(γH1)(x′, t)| ≤ A
Tr · kΠ(·, T)k∞,Qn+1(0, Tκ)· |Qn+1(0, Tκ)|1/p′
· k∂tf((x′,0, t) +·)kp,Qn+1(0, Tκ)
by (2.2) and H¨older’s inequality; hence
(2.4) ≤c∗·T−|κ|·(1−1/p′)−κj· k∂tf((x′,0, t) +·)kp,Qn+1(0, Tκ)
by the kernel-estimate (1.2). Now observe that k∂tf((x′,0, t) +·)kpp,Qn+1(0, Tκ)=
= Z Tκn
0
k∂tf(x′+·, yn, t+·)kp
p,Qˇn+1(0, Tκ)dyn, which easily implies via Fubini’s theorem
(2.5)
Z
. . . Z
Qn−1(λ)×( 0,T)
k∂tf((x′,0, t) +·)kpp,Qn+1(0, Tκ)dx′dt
1/p
≤
≤ |Qˇn+1(0, Tκ)|1/pk∂tfkp,Qn((−λ1′,0),(λ1′+Tκ′,Tκn))×(0,2T). Hence, by the last inequality, (2.4) and since|Qˇn+1(0, Tκ)|=T|κ|−12 andκj = 12:
r.h. side in (2.3)
≤c∗·h·T−12(1+p1)· k∂tfk
p, Qn((−λ1′,0),(λ1′+Tκ′,T1/2))×(0,2T)
so that, abbreviatingρ=ρ(p) := 12(1−p1),
|γH1|
L0,ρp (Qn−1(λ)×(0,T))≤
≤c∗·T−21(1+1p) Z T
0
h−1+p(1−ρ)dh
!1/p
k∂tfkp, Qn(a, b)×(0,2T)
witha:= (−λ1′,0) and b:= (λ1′+Tκ′, T1/2) ; now 1−ρ= 12(1 +1p) and theT factors in the last inequality cancelled, as desired.
Let us turn our attention toH2(i): trivially, forh≤T,
(2.6) k∆t,h(γH2(i))kp,Qn−1(λ)×(0,T−h) ≤ 2· kγH2(i)kp,Qn−1(λ)×(0,T); furthermore, using the kernel estimate (1.3) (withs= 0), we get
(2.7) |γH2(i)(x′, t)| ≤
≤c∗· Z h
0
v−(1+|κ|+εκn)+12 Z
. . . Z
Qn+1(0, vκ)
ynε· |∂ilif((x′,0, t) +y)|dy dv;
we now represent the integrand as
{v−p1′(1+|κ|)+12(ρ−ε·κn)} · {v−1p(1+|κ|−21)+12(ρ−εκn)·yεn· |∂ilif((x′0, t) +y)| } (note that 1/2 =ρ+ 1/2p) ; we chooseε∈(0, ρ/κn); H¨older’s inequality (withp′, p) iny-v space then yields
(2.8) l.h.s. in (2.7)≤c∗· Z h
0 v−1+p
′
2(ρ−ε·κn)dv
!1/p′
·I1/p
with I:=
Z h
0
Z . . .
Z
Qn+1(0, vκ)
v−(1+|κ|−12)+p2(ρ−ε·κn)·ynεp· |∂ilif((x′,0, t) +y)|pdy dv ,
where in the first integral we took into account that|Qn+1(0, vκ)|=v|κ|; the first integral is clearly proportional toh12(ρ−ε·κn). Thus, after a computation as in (2.5), we get
(2.9) kγH2(i)kp,Qn−1(λ)×(0,T)≤c∗·h12(ρ−ε·κn)·I˜1/p with
I˜:=
Z h
0
v−(1+|κ|−12)+p2(ρ−ε·κn)|Qˇn+1(0, vκ)|
Z . . .
Z
Qn+1(a, b(v))
znε·p· |∂ilif(z)|pdz dv ,
where a:= (−λ1′,0,0), b(v) := (λ1′+vκ′, vκn, T +v); since b(v)≤b(h), we can continue
I˜≤ Z h
0
v−1+p2(ρ−ε·κn)dv Z
. . . Z
Qn+1(a, b(h))
znε·p· |∂ilif(z)|pdz
≤ c∗·h(ρ−ε·κn)·p/2 Z hκn
0
znε·p·ϕ(zn)dzn withϕ(zn) :=k∂ilif(·, zn,·)kp
p, Qn−1(−λ1′, λ1′+Tκ′)×(0,2T) by Fubini’s theorem and sinceh≤T; consequently, by (2.6), (2.9) and the last line
(2.10) |γH2|p
L0,ρp (Qn−1(λ)×( 0,T))≤
≤c∗· Z T
0
h−(1+p·ε·κn) Z hκn
0
zε·pn ·ϕ(zn)dzndh
and by the version of Hardy’s inequality from Lemma, ( i) in the Appendix
≤c∗·(p·ε·κn)−1· Z Tκn
0
ϕ(zn)dzn
=c∗·(p·ε·κn)−1· k∂ilifkp
p, Qn((−λ1′,0),(λ1′+Tκ′,T1/2))×(0,2T) , which is the desired result forH2(i).
Finally, let us turn to H3(i); we again use (2.3) and observe that the correct expression for∂t(γH3(i)) is obtained just by replacingKi (in the definition ofH3(i)) by∂n+1Ki (integrate by parts); after estimating|∂n+1Ki| according to (1.3), we arrive at
(2.11) |∂t(γH3(i))(x′, t)| ≤
≤ c∗· Z T
h
v−(1+|κ|+12+ε·κn) Z
. . . Z
Qn+1(0, vκ)
ynε· |∂ilif((x′0, t) +y)|dy dv (cp. (2.7); here the v-exponent is smaller by one, since ∂n+1Ki entails (in (1.3)) the additional factorv−1); in the last integral we write the integrand in the form
{v−p1′(1+|κ|)−(1−ρ−δ)
} · {v−p1(1+|κ|−12)−(εκn+δ)·ynε· |∂ilif(· · ·)|}
(note that−12 = 2p1 +ρ−1), where we introducedδ∈(0,1−ρ). Now apply H¨older’s inequality (withp′, p) iny-v space and get
r.h.s. in (2.11)≤c∗· Z T
h
v−1−p′·(1−ρ−δ)dv
!1/p′
·J1/p with
J :=
Z T
h
v−(1+|κ|−12)−p(ε·κn+δ) Z
. . . Z
Qn+1(0, vκ)
ynεp· |∂ilif((x′,0, t) +y)|pdy dv; proceeding as in the argument leading from (2.8) to (2.9), the last estimate allows us to conclude
k∂t(γH3(i))kp, Qn−1(λ)×(0,T)≤
≤c∗·h−(1−ρ−δ)· Z T
h
v−1−p(ε·κn+δ) Z vκn
0
znε·p·ϕ(zn)dzndv
!1/p
withϕ(·) as before (sincev≤T); by (2.3)
|γH3(i)|p
L0,ρp (Qn−1(λ)×(0,T))
≤c∗· Z T
0
h−1+pδ Z T
h
v−1−p·(ε·κn+δ) Z vκn
0
znε·p·ϕ(zn)dzndv dh
≤c∗·(p·δ)−1· Z T
0
v−1−p·ε·κn Z vκn
0
zεpn ·ϕ(zn)dzndv
by Appendix, Lemma (ii); now we may continue as after (2.10) and the desired result forH3(i) follows.
Thus (2.1) is proved for allT ≤T0(λ) =λ2.
Appendix.
We note a version of Hardy’s inequality.
Lemma. Suppose thatf ∈L1(0, Tγ)is nonnegative, 0< T ≤ ∞;ε, γ >0. Then (i) RT
0 x−1−ε·γ Rxγ
0 yε·f(y)dy dx ≤ (γ·ε)−1 RTγ
0 f(y)dy, (ii) RT
0 x−1+ε·γ RTγ
xγ y−ε·f(y)dy dx ≤ (γ·ε)−1 RTγ
0 f(y)dy.
Proof: These inequalities are proved in BESOV/ IL’IN/ NIKOL’SKII [B/I/N, 2.15, p. 28] (even in a more general form) forT =∞. ForT finite they follow easily by applying the version forT =∞to the extension by zero off to IR+.
References
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University of Bayreuth, Faculty of Mathematics and Physics, P.O.Box 101251, 8580 Bayreuth, Federal Republic of Germany
(Received August 6, 1990)